This is a old problem but a good one. It's been making its rounds through work here. I work with engineers and computer science people - and many of them can't get it right.<br><br>It's the old Monty Hall conundrum:<br><br>Basically you are given 3 doors. You get to choose one of the doors that contains the car. The other two contain goats. (It is assumed you want the car, not a goat).<br><br>You make your choice. But before revealing the choice, Monty opens one of the doors you did not choose to reveal a goat.(No matter what door you have chosen, Monty will always open another door to reveal a goat) He then gives you the option to change your selection to the other door before opening it to reveal your prize. If you want the car, should you switch? If so, why?<br><br>(no cheating with Google )<br><br><P ID="edit"><FONT SIZE=-1><EM>Edited by hayesk on 09/15/04 09:31 AM (server time).</EM></FONT></P>

No need to switch. You have two doors and one has a car. 50% chance of getting it right by switching or not switching. This is a problem for people who cannot do a simple statistical analysis on a problem. The hardest part in this one is that one has to completely ignore what has gone before. You have already played for the three doors. That is past history and has nothing to do with the choice you have in front of you. Think that way and then it is obvious.<br><br>nb:Now since Monty knows if you have picked correctly (since he had to since he knew where at least one goat was) you could look into his eyes and try to fathom whether to switch or not. Give him a whole pitiful story about all the orphans you need to transport and if he tears up and looks at one door I would go with that. Monty was a good show because I think he body languaged the best response from the contestant. The greedy seemed to always crash and burn and the poor widower got the door.<br><br><br><br><br><br><br>luciferase is a four nineteener

Heh heh, you are incorrect. Your first choice has everything to do which doors Monty picked - so it does matter in your odds if switching. I will give a hint later that illustrates it well.<br><br>BTW, Many a university math professor has been stumped by this, only to later admit they were wrong.<br><br>

Two hints:<br><br>Hint 1. There is no "trick" or anything you need to assume other than Monty will always show you a losing door - i.e. he knows where the prize is.<br><br>Hint 2. If there were 10000 doors, you choose a door, and then Monty revealed 9998 doors with goats in them, would you be confident that your first choice has the car, or the one door (out of 9999 doors) delibritely left closed for you to switch to (or not) by Monty? The key here is that Monty skews the odds by deliberately showing you goats.<br><br><br>

Well I will have to wait for your evidence. Your second example is a good one of your flawed logic. Sure, Monty has removed all doors but the one you hold and one other. Monty knows whether you have one or lost. But you and you alone make the choice and you have no other evidence. It is a 50 50 of choice.<br><br>Monty has played a mind game on the contestant. He starts with a 1/3 chance with three doors. You pick. Then Monty cancels that game and makes up another one. Nothing prior to the last choice has anything to do with the odds on that choice. As stated Monty would have said the exact same thing to you if you had selected a goat or a car the first time. You would have still had the same choice, stand pat or change. Thus 50%.<br><br>If I am wrong then I promise to stop being a jackass 90% of the time. I would go to 50% jackass. (Hold the applause I haven't lost yet.)<br><br><br><br><br><br>luciferase is a four nineteener

Heh heh. Ok, here it is. (spoiler alert)<br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br>The odds of switching are 2/3 for one prize and three doors - always. There are a few ways to explain it:<br><br>Explanation 1. My logic is not flawed because you asssume (even in my 10000 door example) that the two remaining closed doors have equal probability of having the car. Only the first choice of door is truly random and equal probability. The second closed door (one left out of 2 or out of 9999) is specifically chosen by Monty.<br><br>Scenario 1 - If I guessed wrong on my first door chosen(which is n-1/n probability), the second door has a 100% probability, since that is the only door Monty could have left closed - he has no other choice but to leave the winning door closed.<br>Scenario 2 - If I guessed correctly the first door, which only has a 1/n chance, then Monty can choose any of the n-1 doors, and switching would lose.<br><br>Scenario 2 only has a 1/n chance of occurring, but the first scenario giving a 100% chance of winning by switching, occurs n-1/n of the time. Thus, since you don't know where the car is, it's better odds to always switch because n-1 times out of n, you picked the wrong door first giving you a 100% chance of winning by switching to the other door.<br><br>Explanation 2. You choose a door. You know Monty is going to show you a goat anyway, so it doesn't affect your decision to stay or switch, since no new information is given to you by Monty opening a door. Your odds of picking the car the first time are 1/3. Therefore, picking and staying is 1/3, so picking and switching must be 2/3. (1 - odds of picking correctly on first choice)<br><br>Explanation 3. Count it.<br><br>Pick any door, switch or stay - 3 doors * 2 choices (switch or stay) = 6 choices. Let's say prize is in B. We don't know, Monty does.<br>Pick A, Monty must show C, stay and lose.<br>Pick A, Monty shows C, switch and win.<br>Pick B, Monty can show A or C, stay and win.<br>Pick B, Monty can show A or C, switch and lose.<br>Pick C, Monty shows A, stay and lose.<br>Pick C, Monty shows A, switch and win.<br><br>2/3 of the wins are by switching. 1/3 is by staying.<br><br>A coworker wrote a computer program to simulate this too. 2/3 times winning was by switching to the other door.<br><br>

Forgot to address your reasoning:<br><br><blockquote><font size=1>In reply to:</font><hr><p>Monty has played a mind game on the contestant. He starts with a 1/3 chance with three doors. You pick. Then Monty cancels that game and makes up another one. Nothing prior to the last choice has anything to do with the odds on that choice.<p><hr></blockquote><p><br>This is where you goofed. The first choice has everything to do with it. Your first choice determines whether Monty has one door to choose from to open, or two doors. 1/3 of the time, Monty can choose either door. 2/3 of the tiem, Monty must choose a specific door. Again, think of the 10000 example: do you really think your odds of choosing your first door correctly, which you know to be 1/10000 is the same as the one door Monty purposefully left for you to switch to.<br><br><br>

You should switch. Here's my reasoning:<br><br>You have Goat A, Goat B, and Car:<br>If you choose Car, then a switch will give you a Goat (A or B dpending what he show).<br>If you choose Goat A, then a switch will give you the Car (because he will reveal Goat B).<br>If you choose Goat B, then a switch will give you the Car (because he will reveal Goat A).<br><br>Therefore 2 of the 3 scenarios will gve you the car if you switch (or a 66% chance).<br><br>-Matt<br><br>damn, you posted the solution when I was writing my reasoning<P ID="edit"><FONT SIZE=-1><EM>Edited by LoveTheBomb on 09/15/04 11:56 AM (server time).</EM></FONT></P>

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